Cho he phuong trinh
mx + y=1
x-my=2
Biet (xo;yo) la nghiem cua he. Xac dinh m sao cho xo+yo=1
Giúp em với mấy anh chị ơi!!!!!!!!!!!!!!!!!!
Cho he phuong trinh sau:
\(\hept{\begin{cases}\left(m+1\right)x+my=2m-1\\mx-y=m^2-2\end{cases}}\)
Tim m de he phuong trinh co nghiem duy nhat (x;y) thoa man P= xy dat gia tri lon nhat.
Cho he phuong trình
mx + y =1
x –my =2
Biet (x0; y0) laø nghiem cua he. Xac dinh m sao cho x0 + y0 = 1
theo mx+y=1 =>y=1-mx
ta có x-my=2=>x-m(1-mx)=2=>x-m+m^2x=2=>x(1+m^2)=2+m=>x=(2+m)/(1+m^2)
thay vô tim y=(2m^2-m)/((1+m^2)/m)
thay vô x+y=1 tim ra m
Cho he phuong trinh
mx + y =1
x – my =2
Biet (x0; y0) la nghiem cua he. Xac dinh m sao cho x0 + y0 =1
Giúp em với !!!!!!!!!!!!!!!!!!
Cho he phuong trinh: x-my=0
mx-y=m+1 (m la tham so)
a Giai va bien luan he phuong trinh tren
b Tim m de hpt co nghiem duy nhat thoa man
1 M(x,y) cach deu 2 truc toa do
2 P(2,4) va Q(-2,-6) doi xung qua M(x,y)
cho he phuong trinh ( m-1)x -my = 3m-1 va 2x -y = m=5
a. tim m de he co nghiem duy nhat ma S = x2 + y2 dat gia tri nho nhat
\(\hept{\begin{cases}x+my=m+1\left(1\right)\\mx+y=3m-1\left(2\right)\end{cases}}\)
tim m de he phuong trinh co nghiem duy nhat (x;y) sao cho x;y co gia tri nho nhat
cho he phuong trinh 3x-y=2m+3 va x+2y=3m+1 tim m de he phuong trinh co 2 nghiem x y thoa man x^2+y^2=5
\(\hept{\begin{cases}3x-y=2m+3\\x+2y=3m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}6x-2y=4m+6\\x+2y=3m+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=m+1\\y=m\end{cases}}\)khi đó: \(^{x^2+y^2=5\Leftrightarrow2m^2+2m+1=5\Leftrightarrow2m^2+2m-4=0\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}}\)
tinh gia tri cac bieu thuc sau:3x^4+5x^2y^2+2y^4+2y^2biet rang x^2+y^2=1
\(=3x^4+3x^2y^2+2x^2y^2+2y^4+2y^2\)
\(=\left(3x^2+2y^2\right)\left(x^2+y^2\right)+2y^2\)
\(=3x^2+2y^2+2y^2\)
\(=3x^2+4y^2\)
Cho he phuong trinh
\(\left(a+1\right)x-ay=5\) (1)
\(x+ay=a^2+4a\) (2)
Tim gia tri cua a thuoc Z sao cho he phuong trinh co nghiem (x;y) voi x, y thuoc Z
Lấy (1) cộng (2), ta có:
\(\left(2a+1\right)x=a^2+4a+5\)\(\Rightarrow x=\dfrac{a^2+4a+5}{2a+1}\)
Thay vào (1): \(\dfrac{\left(a^2+4a+5\right)\left(a+1\right)-10a-5}{2a+1}.\dfrac{1}{a}\)\(=\dfrac{a^3+5a^2-a}{2a+1}.\dfrac{1}{a}=\dfrac{a^2+5a-1}{2a+1}\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}a^2+4a+5⋮2a+1\\a^2+5a-1⋮2a+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a+2\right)+2a+5⋮2a+1\\a^2+2a+3a-1⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\a+2⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\3⋮2a+1\end{matrix}\right.\)\(\Rightarrow2a+1\in\left\{\pm1\right\}\)\(\Rightarrow a\in\left\{-1;0\right\}\)
Vậy với a=-1;0 thì hpt có nghiệm (x;y) với x,y thuộc Z.